where $A_{n_j}=f^{-1}[\frac{j}{2^n},\frac{j+1}{2^n})$ for $j\neq n2^n$ and $A_{n_j}=f^{-1}[n,\infty)$ for $j=n2^n$.

I see intuitively (by drawing pictures) why this is a uniform approximation, and where the second equality comes from. However I can't see how to prove these rigorously in a clean way. Does anyone have a clean and insightful proof of (a) the second equality and (b) that the $f_n$ uniformly approximate $f$?

Thanks for the pointers - I've edited the question so that the expression is now right. So you are saying that the $\wedge n$ part of the formula is unnecessary then? Presumably it is necessary if $f$ is unbounded, in which case this still works, right?
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Edward HughesMay 24 '12 at 13:21